Question

Let \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \] Then, \( \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} \) is equal to:

• A. 1
• B. 0
• C. \( \infty \)
• D. \( -1 \)

Hint:

For limits involving exponential functions and trigonometric functions, consider using L'Hopital's Rule to simplify the expression when the limit results in an indeterminate form.

Answer 1

The Correct Option is A

Step 1: Analyze the Given Function \( f(x) \)

We are given the function: \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) - \tan \left( \frac{x}{2^{r+2}} \right)}{1} \right) \] This expression simplifies to: \[ f(x) = \tan x \]

Step 2: Set Up the Limit Calculation

Next, we calculate the limit: \[ \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} \]

Step 3: Apply L'Hopital's Rule

Since this is an indeterminate form of type \( \frac{0}{0} \), we apply L'Hopital's Rule, which involves differentiating the numerator and denominator: \[ \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} \] Differentiating the numerator: \[ \frac{d}{dx} \left( e^x - e^{\tan x} \right) = e^x - e^{\tan x} \cdot \sec^2 x \] Differentiating the denominator: \[ \frac{d}{dx} (x - \tan x) = 1 - \sec^2 x \] Substituting at \( x = 0 \): \[ \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} = 1 \]

Final Answer: 1

Answer 2

Step 1: Understanding the given function.
We are given the function:
\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \]
We need to find:
\[ \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)}. \]

Step 2: Simplify the summation term.
We know that: \[ \tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}. \] However, our numerator and denominator are different, so let’s simplify the given expression:
\[ \frac{\tan\theta + \tan^3\theta}{1 - \tan^2\theta} = \tan(2\theta), \] because using the tangent double-angle identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}. \] Thus the given expression is equivalent to: \[ \frac{\tan\theta + \tan^3\theta}{1 - \tan^2\theta} = \tan(2\theta) \text{ when small-angle approximation holds.} \]

Step 3: Simplify the summation expression for small angles.
For very small \( x \), \( \tan x \approx x \). Thus:
\[ \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \approx \frac{\frac{x}{2^{r+1}}}{1} = \frac{x}{2^{r+1}}. \] Hence, \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \frac{x}{2^{r+1}} = x \lim_{n \to \infty} \sum_{r=0}^{n} \frac{1}{2^{r+1}}. \] The sum of the infinite geometric series is: \[ \sum_{r=0}^{\infty} \frac{1}{2^{r+1}} = 1. \] Thus: \[ f(x) = x. \]

Step 4: Evaluate the given limit.
We now substitute \( f(x) = x \) into the required expression: \[ \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^x - e^x}{x - x}. \] This is an indeterminate form \( \frac{0}{0} \), but since \( f(x) = x \), both numerator and denominator vanish in the same way.

By applying L'Hôpital's Rule (or recognizing that for small deviations, the exponential behaves linearly), the limit evaluates to: \[ \boxed{1}. \]

Final Answer:
\[ 1 \]